Answer:
Mass = 76.176 g
Explanation:
Given data:
Mass of lead(II) chloride produced = 62.9 g
Mass of lead(II) nitrate used = ?
Solution:
Chemical equation:
Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃
Number of moles of lead(II) chloride:
Number of moles = mass/molar mass
Number of moles = 62.9 g/ 278.1 g/mol
Number of moles = 0.23 mol
Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:
PbCl₂ : Pb(NO₃)₂
1 : 1
0.23 : 0.23
Mass of Pb(NO₃)₂:
Mass = number of moles × molar mass
Mass = 0.23 mol × 331.2 g/mol
Mass = 76.176 g
240 g NaOH
<em>Step 1</em>. Calculate the moles of NaOH
Moles of NaOH = 0.750 L solution × (8 mol NaOH/1 L solution) = 6 mol NaOH
Step 2. Calculate the mass of NaOH
Mass of NaOH = 6 mol NaOH × (40 g NaOH/1 mol NaOH) = 240 g NaOH
Answer:
Explanation:
They play a very important part. The geometry is not a straight line. It is an angle over 90 which means that the molecule has the same general shape as a boomerang. The two hydrogens and the 2 lone electron pairs try to get away as far as possible from each other. The actual shape results in a tetrahedron shape. But the two hydrogens and 1 oxygen actually look like the aforementioned boomerang.
Answer:
The answer should be 40.3044 g/mol
Answer:
The average mass of the gold coin reported to five significant figures, even though you had to divide by "3" to obtain the average.
Explanation:
Given that,
The mass of a silver liberty was measured three times and each measurement was made to five digits.
The mass values are,
The average mass of the gold coin is 12.512 g.
We know that,
Significant figures :
Significant figures is explained the nonzero, leading zero and zeros between two nonzero digits.
For example,
The digits is 0.003405
In the digit, 345 = nonzero
0.00 = leading zero
The average mass of the gold coin reported to five significant figures because average mass is 12,512. All digits are nonzero.
Nonzero digit are significant.
Hence, The average mass of the gold coin reported to five significant figures, even though you had to divide by "3" to obtain the average.
